I'm still not sure how to find the melting of the ice, do i need to use latent heat constant?
ahhh so add the two values together and then multiply by the area*the density?
1.if i raise the temp of the ice to the melting pt. it would need to be at its boiling pt 100 degrees celcius. heat rises
2. now must melt the ice at 0 degrees celcius
I'm sorry I really don't know what I'm doing just learned this for the first time today along with four other physics chapter...
Homework Statement
A woman finds the front windshield of her car covered with ice at -12.8°C. The ice has a thickness of 4.60 10-4 m, and the windshield has an area of 1.25 m2. The density of ice is 917 kg/m3. How much heat is required to melt the ice?
Homework Equations
Q=mL or change...
first i would take the celcius and subtract 32 degrees since its above the farenheight freezing pt. and then convert to farenheight by multiplying by (9/5)....right....and in the other part i would add 32 degrees since its below the zero pt??
Homework Statement
On the moon the surface temperature ranges from 371 K during the day to 1.09 102 K at night.
(a) What are these temperatures on the Celsius scale?
(b) What are these temperatures on the Fahrenheit scale?
Homework Equations
T=Tc+273.15 T=Kelvin temperature Tc=Celcius...
So I'm trying to learn how to do it using the angular velocity way. I'm not familiar with this method and I was hoping if you could check my work.
a = omega^2 r
So a is proportional to r.
a(r) = a(@ r') (r / r')
plug the numbers in and i get
a(.050)=118(.030)(.050/.030)
=118<<i don't...
I haven't learned angular velocity yet so I'm not sure how to relate the two. I was trying to use the formula for centrepitual acceleration A=v^2/r
I then set up the equation for the first distance from the center of the CD and then set up a second equation for the second distance from the...
Homework Statement
A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 118 m/s2. What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?
Homework Equations...
V_y=75m/s*sin(66.4)
then I use the formula (V_y*t-9.80*t^2)/2
then subtract that answer from 11m right?<<<am I'm on the right track I just keep getting a very large numbers for answers